Given a sorted array of integers, find the number of occurrences of a given target value. Your algorithm’s runtime complexity must be in the order of O(log n). If the target is not found in the array, return 0
**Example : ** Given [5, 7, 7, 8, 8, 10] and target value 8, return 2.
思路:
做两次binary search,一次找到最左边的index,一次找到最右边的index
Solution:
Time: O(logn) Space: O(1)
public class Solution { // DO NOT MODIFY THE LIST. IT IS READ ONLY // 0 1 2 3 4 5 // [5, 7, 7, 8, 8, 10] public int findCount(final List<Integer> A, int B) { int l = 0; int left = 0; int right = A.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (A.get(mid) >= B) { right = mid - 1; } else { left = mid + 1; l = left; } } // System.out.println(l); left = 0; right = A.size() - 1; while (left <= right) { int mid = left + (right - left) / 2; if (A.get(mid) <= B) { left = mid + 1; } else { right = mid - 1; } } // System.out.println(right); int result = right - l + 1; return result > 0 ? result : 0; } }